Let $M$ be a $n$-dimensional topological manifold, and $U \subseteq M$ be an open set such that there is a homeomorphism $\phi: U \to B$ where $B$ is the $n$-dimensional open unit ball.
Question: Let $x \in M$ be a point in $\partial U$ (assume such $x$ exists), is there a point $y \in \partial B$ and a homeomorphism $\tilde \phi: U \cup \{x\} \to B \cup \{y\}$ such that $\tilde \phi|_U = \phi$?
The statement is definitely false if $M$ is an arbitrary topological space. However, I wonder whether it is still false when $M$ is a topological manifold. I encountered this question when dealing with a problem on smooth manifolds, and I'm not familiar at all with theories of topological manifolds.
Given $U = \Bbb C \setminus \Bbb R_-$, it is known that $U$ is homeomorphic to the unit disc $\Delta$ (for example, by Poincaré's theorem, they are biholomorphic).
Let $-1 \in \partial U$. Then $U \cup\{-1\}$ is not simply connected, while for any $y \in \partial \Delta$, $\Delta \cup \{y\}$ is. It follows that they are not homeomorphic, and the homeomorphism $U \to \Delta$ cannot be extended into a homeomorphism $U\cup\{-1\} \to \Delta \cup\{y\}$.